First you need to identify the hidden pair in the bottom left 3x3 box. Look at the pair {4, 9} appearing in row 7 and in row 9 -- the only way {4, 9} can fit in the bottom left 3x3 box is if it goes in r8c1 & r8c3.

Now look at column 3, in the top left 3x3 box. The only values possible at r1c3 are {4, 8}. The only values possible at r2c3 are {4, 8, 9}. So there's a triplet {4, 8, 9} lying in column 3, at r1c3, r2c3, and r8c3.

Now looking across row 5, it appears that the possibilities for r5c3 are {1, 4, 8} -- but the "4" and the "8" are out because of the triplet. dcb

There is a quadruple of 1,2,3,7 in Box No. 7 ( cells r7c2, r9c1, r9c2, r9c3 ). This then leaves you with a pair of 4,9 in r8c1 and r8c3.

You then have a triplet of 4,8,9 in Column 3 ( cells r1c3, r2c3 and r8c3). Therefore you can eliminate all other 4,8,9s from cells in Column 3. Since the only possibilities for Cell No. r5c3 are 1,4,8 this then leaves Nr 1 as only possibility in that cell.